Question

Let \( X \) be a random variable with the distribution function \[ F(x) = \begin{cases} 0, & x < 0, \\ \frac{1}{4} + \frac{4x - x^2}{8}, & 0 \leq x < 2, \\ 1, & x \geq 2. \end{cases} \] Then \[ P(X = 0) + P(X = 1.5) + P(X = 2) + P(X \geq 1) \] equals 
 

• A. \( \frac{3}{8} \)
• B. \( \frac{5}{8} \)
• C. \( \frac{7}{8} \)
• D. 1

Hint:

When working with distribution functions, be careful with jumps at discrete values and ensure to calculate the probability mass at each point separately.

Answer 1

The Correct Option is C

Step 1: Understand the distribution function. 
To find \( P(X = 0) \), \( P(X = 1.5) \), and so on, we first calculate the individual probabilities: \[ P(X = 0) = F(0) - F(0-) = \frac{1}{4} \text{(as the value jumps at \( x = 0 \))}. \] For \( P(X = 1.5) \): \[ P(X = 1.5) = F(1.5) - F(1.5-) = \frac{1}{4} + \frac{4(1.5) - (1.5)^2}{8} - \left(\frac{1}{4} + \frac{4(1) - (1)^2}{8}\right). \] Simplifying this gives: \[ P(X = 1.5) = \frac{7}{8} - \frac{5}{8} = \frac{1}{4}. \] For \( P(X = 2) \), note that \( P(X = 2) = 0 \) as the function jumps at \( x = 2 \).
 

Step 2: Calculate total probability. 
Now, calculate \( P(X \geq 1) \): \[ P(X \geq 1) = 1 - F(1-) = 1 - \left(\frac{1}{4} + \frac{4(1) - (1)^2}{8}\right) = 1 - \frac{5}{8} = \frac{3}{8}. \] Now, summing the probabilities: \[ P(X = 0) + P(X = 1.5) + P(X = 2) + P(X \geq 1) = \frac{1}{4} + \frac{1}{4} + 0 + \frac{3}{8} = \frac{7}{8}. \]

Step 3: Conclusion. 
The correct answer is (C) \( \frac{7}{8} \).